Practice Set - Week 3

Author

SOPHAS

Published

16-Apr-2025

Q1. Define and explain the following terms:

Answer

---
title: Coin Toss (Statistical Experiment) 
---
flowchart TB
  A((Coin is tossed))
  A -->|Random Phenomenon| B{Outcome}
  B --> |Random Event|C(Heads)
  B --> |Random Event|D(Tails)
  F("Sample Space: {Heads,Tails}")

Q2. What is probability? Discuss its importance.

Answer

Probability is a fundamental concept in statistics, playing a crucial role in understanding and quantifying uncertainty in various phenomena.

It’s importance lies in its ability to support informed decision-making. Probability allows statisticians and researchers to make predictions about future events, draw inferences about a population based on sample data, and assess the reliability of those inferences. It also forms the foundation for many statistical methods, including hypothesis testing, confidence intervals, and regression analysis. Overall, probability is essential for modeling randomness, managing risk, and making sound conclusions in the presence of uncertainty.

Q3. Differentiate between the following pairs of concepts:

Answer

  1. Elementary Event Compound Event
    An event that consists of a single outcome. An event that consists of two or more outcomes.
    Example: Getting a 4 when a die is rolled. Example: Getting an even number (2, 4, or 6) on a die.


  2. Mutually Exclusive Events Overlapping Events
    Events that cannot occur at the same time. Events that can occur at the same time.
    The occurrence of one event rules out the occurrence of the other. The events share at least one common outcome.
    Example: Getting a head or a tail in a single coin toss. Example: Drawing a red card or a king from a deck (a red king belongs to both).


  3. Sample Space Sample Point
    The set of all possible outcomes of an experiment. A single outcome from the sample space.
    Denoted by ‘S’. Each element of the sample space.
    Example: For tossing a coin twice, S = {HH, HT, TH, TT}. Example: HT is a sample point.

Q4. Three unbiased coins are tossed. Calculate the probability of:

Answer

The sample space for 3 unbiased coin tosses is \(2^3 = 8\) outcomes, since each coin has 2 possible results (Head or Tail):

flowchart TB
    A[Start] --> B1[H]
    A --> B2[T]

    B1 --> C1[HH]
    B1 --> C2[HT]
    B2 --> C3[TH]
    B2 --> C4[TT]

    C1 --> D1[HHH]
    C1 --> D2[HHT]
    C2 --> D3[HTH]
    C2 --> D4[HTT]
    C3 --> D5[THH]
    C3 --> D6[THT]
    C4 --> D7[TTH]
    C4 --> D8[TTT]

Each coin has two outcomes: Head \(\small{(H = 1)}\) or Tail \(\small{(T = 0)}\)

Binomial Distribution

In a coin toss experiment, each toss is a Bernoulli trial (success = heads with probability \(\small{p}\) , failure = tails with probability \(\small{1-p}\)).

The binomial distribution models the probability of getting exactly \(\small{k}\) successes (e.g., heads) in \(\small{n}\) independent tosses.

  • The probability mass function (PMF) gives the probability of getting exactly \(\small{k}\) heads: \(\small{P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}}\)

  • The cumulative distribution function (CDF) gives the probability of getting at most \(\small{k}\) heads: \(\small{P(X \le k) = \sum_{i=0}^{k} \binom{n}{i} p^i (1 - p)^{n - i}}\)

This will be covered further in the binomial distribution section.

Julia uses the // operator to create exact rational numbers as ratios of integers.

# Create PMF and CDF functions

my_pmf(k::Int64; n::Int64=3, p::Rational=1//2) = binomial(n, k) * p^k * (1 - p)^(n - k)

my_cdf(k::Int64; n::Int64=3, p::Rational=1//2) = sum(binomial(n, i) * p^i * (1 - p)^(n - i) for i in 0:k)
my_cdf (generic function with 1 method)

In Julia, functions can include keyword arguments that provide default values.

For example, in the function my_pmf(k::Int64; n::Int64=3, p::Rational=1//2), n and p are keyword arguments with default values of 3 and 1/2, respectively.

These can be overridden by specifying them explicitly when calling the function, like my_pmf(2; n=5, p=2//3)

all_heads = my_pmf(3)

println("P(X = 3): $all_heads")
P(X = 3): 1//8
two_heads = my_pmf(2)

println("P(X = 3): $two_heads")
P(X = 3): 3//8
one_head = my_pmf(1)

println("P(X = 3): $one_head")
P(X = 3): 3//8
atleast_1_head = 1 - my_cdf(0)

println("P(X ≥ 1): $atleast_1_head")
P(X ≥ 1): 7//8

In a discrete distribution, \(\small{P(X < k) = P(X \le k-1)}\) because there are no values between the integers.

atleast_2_heads = 1 - my_cdf(1)

println("P(X ≥ 2): $atleast_2_heads")
P(X ≥ 2): 1//2
all_tails = my_pmf(0)

println("P(X = 0): $all_tails")
P(X = 0): 1//8
Event Favorable Outcomes Probability
All heads (H, H, H) \(\small{\dfrac{1}{8}}\)
Two heads (H, H, T), (H, T, H), (T, H, H) \(\small{\dfrac{3}{8}}\)
One head (H, T, T), (T, H, T), (T, T, H) \(\small{\dfrac{3}{8}}\)
At least one head All outcomes except (T, T, T) \(\small{\dfrac{7}{8}}\)
At least two heads (H, H, T), (H, T, H), (T, H, H), (H, H, H) \(\small{\dfrac{4}{8} = \dfrac{1}{2}}\)
All tails (T, T, T) \(\small{\dfrac{1}{8}}\)

Q5. A card is drawn from a well-shuffled deck of 52 cards. Determine the probability of drawing a card that is neither a heart nor a king.

Answer

We need to find the probability of drawing a card that is neither a heart nor a king, i.e., \(\small{P(\text{Neither a heart nor a king})}\):

# Given data
deck = 52
kings = 4
hearts = 13
king_hearts = 1

# Cards that are neither a heart or a king
heart_nor_king = deck - (kings + hearts - king_hearts)

# Probability
p = heart_nor_king // deck

println("The P(neither a heart nor a king): $p")
The P(neither a heart nor a king): 9//13

Q6. In a single throw of two dice, find the probability of:

Answer

The sample space for a single throw of two dice:

\[ \small \begin{aligned} S = \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), \\ (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), \\ (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), \\ (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), \\ (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), \\ (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\} \end{aligned} \]

# Total possible outcomes when rolling two dice
total_outcomes = 6 * 6

# Count outcomes where the sum is 11
favorable_11 = 0

for die1 in 1:6, die2 in 1:6
    if die1 + die2 == 11
        favorable_11 += 1
    end
end

# Probability
p_11 = favorable_11 // total_outcomes

println("P(Total = 11): $p_11")
P(Total = 11): 1//18
# Total possible outcomes when rolling two dice
total_outcomes = 6 * 6

# Count outcomes where the sum is 8 or 11
favorable_8_or_11 = 0

for die1 in 1:6, die2 in 1:6
    if die1 + die2 == 8 || die1 + die2 == 11
        favorable_8_or_11 += 1
    end
end

# Probability
p_8_or_11 = favorable_8_or_11 // total_outcomes

println("P(Total = 11): $p_8_or_11")
P(Total = 11): 7//36
# Total possible outcomes when rolling two dice
total_outcomes = 6 * 6

# Count favorable outcomes: same number on both dice 
favorable_same = 0

for die1 in 1:6, die2 in 1:6
    if die1 == die2
        favorable_same += 1
    end
end

# Probability 
p_same = favorable_same // total_outcomes

println("P(Same number on both dice) = $p_same")
P(Same number on both dice) = 1//6
Event Favorable Outcomes Probability
Total of 11 (5,6), (6,5) \(\small{\dfrac{2}{36} = \dfrac{1}{18}}\)
Total of 8 or 11 (2,6), (3,5), (4,4), (5,3), (6,2), (5,6), (6,5) \(\small{\dfrac{7}{36}}\)
Same number on both dice (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) \(\small{\dfrac{6}{36} = \dfrac{1}{6}}\)

Julia provides a variety of control flow constructs, one of which is repeated evaluation using loops. There are two main types: while and for loops.

Q7. Five men in a company of 20 are graduates. If 3 men are picked randomly, what is the probability that:

Answer

all_3 = binomial(5, 3) // binomial(20, 3)

println("P(All three are graduates): $all_3")
P(All three are graduates): 1//114
atleast_1 = 1 - (binomial(15, 3) // binomial(20, 3))

println("P(At least one is a graduate): $atleast_1")
P(At least one is a graduate): 137//228

In Julia, the binomial(n, k) function computes the binomial coefficient:

\(\small{\binom{n}{k} = \dfrac{n!}{k!(n - k)!}}\)

It gives the number of ways to choose k items from n without regard to order — also known as “n choose k”.

Q8. A bag contains 25 balls numbered 1 to 25. Considering an odd number as a ‘success’, and drawing two balls with replacement, find the probability of:

Answer

Each draw is a Bernoulli trial:

Success = odd number = \(\small{13}\) odd numbers in \(\small{1 \text{ to } 25}\), so, \(\small{p = \dfrac{13}{25}}\)

Number of trials: \(\small{n = 2}\)

two_successes = my_pmf(2; n = 2, p = 13//25)

println("P(X = 2): $two_successes")
P(X = 2): 169//625
one_success = my_pmf(1; n = 2, p = 13//25)

println("P(X = 1): $one_success")
P(X = 1): 312//625
atleast_1 = 1 - my_cdf(0; n = 2, p = 13//25)

println("P(X = 1): $atleast_1")
P(X = 1): 481//625
no_successes = my_pmf(0; n = 2, p = 13//25)

println("P(X = 0): $no_successes")
P(X = 0): 144//625
Event Probability
Two successes \(\small{\dfrac{169}{625}}\)
Exactly one success \(\small{\dfrac{312}{625}}\)
At least one success \(\small{\dfrac{481}{625}}\)
No successes \(\small{\dfrac{144}{625}}\)

Q9. A bag contains 5 white and 8 red balls. Two drawings of 3 balls are made:

Find the probability that the first drawing will give 3 white and the second 3 red balls in each case.

Answer

  1. First draw: 3 white out of 5, from 13 total balls: \(\small{P(\text{3 white}) = \dfrac{\binom{5}{3}}{\binom{13}{3}} = \dfrac{10}{286} = \dfrac{5}{143}}\)
three_white = binomial(5, 3) // binomial(13, 3)

println("P(3 white): $three_white")
P(3 white): 5//143
  1. Second draw: 3 red out of 8, from 13 total balls: \(\small{P(\text{3 red}) = \dfrac{\binom{8}{3}}{\binom{13}{3}} = \dfrac{56}{286} = \dfrac{28}{143}}\)
three_red = binomial(8, 3) // binomial(13, 3)

println("P(3 red): $three_red")
P(3 red): 28//143
  1. First draw: 3 white out of 5, from 13 total balls: \(\small{{P(\text{3 white}) = \dfrac{\binom{5}{3}}{\binom{13}{3}} = \dfrac{10}{286} = \dfrac{5}{143}}}\)
three_white = binomial(5, 3) // binomial(13, 3)

println("P(3 white): $three_white")
P(3 white): 5//143
  1. Second draw: 3 red out of 8, from 10 (13-3) total balls: \(\small{P(\text{3 red}) = \dfrac{\binom{8}{3}}{\binom{10}{3}} = \dfrac{56}{120} = \dfrac{7}{15}}\)
three_red = binomial(8, 3) // binomial(10, 3)

println("P(3 red): $three_red")
P(3 red): 7//15
Case First Draw: 3 White Second Draw: 3 Red
With Replacement \(\small{\dfrac{10}{286} = \dfrac{5}{143}}\) \(\small{\dfrac{56}{286} = \dfrac{28}{143}}\)
Without Replacement \(\small{\dfrac{10}{286} = \dfrac{5}{143}}\) \(\small{\dfrac{56}{120} = \dfrac{7}{15}}\)

Q10. Three groups of workers contain 3 men and 1 woman, 2 men and 2 women, and 1 man and 3 women respectively. One worker is selected at random from each group. What is the probability that the selected group consists of 1 man and 2 women?

Answer

Group Composition:

Favorable Cases (1 man, 2 women):

We need exactly one M and two W, so the combinations are:

Selection Type Group A Group B Group C Probability(A * B * C)
M W W \(\small{\dfrac{3}{4}}\) \(\small{\dfrac{2}{4}}\) \(\small{\dfrac{3}{4}}\) \(\small{\dfrac{18}{64}}\)
W M W \(\small{\dfrac{1}{4}}\) \(\small{\dfrac{2}{4}}\) \(\small{\dfrac{3}{4}}\) \(\small{\dfrac{6}{64}}\)
W W M \(\small{\dfrac{1}{4}}\) \(\small{\dfrac{2}{4}}\) \(\small{\dfrac{1}{4}}\) \(\small{\dfrac{2}{64}}\)
Multiplication Rule for Independent Events:

If events \(\small{A}\) and \(\small{B}\) are independent, then \(\small{P(A \cap B) = P(A) \cdot P(B)}\)

Total Probability:

p1 = (3//4) * (2//4) * (3//4)
p2 = (1//4) * (2//4) * (3//4)
p3 = (1//4) * (2//4) * (1//4)

total = p1 + p2 + p3
println("P(1 man and 2 women): $total")
P(1 man and 2 women): 13//32

Q11. What is the probability that a randomly selected leap year will contain 53 Sundays?

Answer

# Total number of possible extra day combinations in a leap year
total_outcomes = 7

# Favorable outcomes: (Saturday, Sunday) and (Sunday, Monday)
favorable_outcomes = 2

# Probability
prob_53_sundays = favorable_outcomes // total_outcomes

println("P(53 Sundays) = $prob_53_sundays")
P(53 Sundays) = 2//7

Q12. A university has to select an examiner from a list of 50 persons, which includes 20 women and 30 men, 10 who know Hindi and 40 who do not, 15 teachers and 35 non-teachers. What is the probability of selecting a Hindi-knowing woman teacher?

Answer

Given:

Probability that a randomly selected person is a hindi knowing woman teacher, assuming each trait is independent since no overlap is specified: \(\small{P(\text{Hindi-knowing} \cap \text{Woman} \cap \text{Teacher}) = \dfrac{10}{50} \times \dfrac{20}{50} \times \dfrac{15}{50} = \dfrac{3000}{125000} = \dfrac{3}{125}}\)

# Total people
total = 50

# Individual probabilities
p_hindi     = 10 // total
p_woman     = 20 // total
p_teacher   = 15 // total

# Final probability assuming independence
p_all_three = p_hindi * p_woman * p_teacher

println("P(Hindi-knowing woman teacher) = $p_all_three")
P(Hindi-knowing woman teacher) = 3//125

Q13. A hospital has two departments for conducting medical tests. Department A conducts 80% of the tests, while Department B conducts 20% of the tests. In Department A, 85 out of 100 tests are accurate. In Department B, 65 out of 100 tests are accurate. What is the probability that a randomly selected accurate test came from Department A?

Answer

Using Bayes’ Theorem, \(\small{P(\text{Dept A} \mid \text{Accurate})}\)

Define Events:

Given Probabilities:

Use Bayes’ Theorem: \(\small{P(A \mid \text{Acc}) = \dfrac{P(\text{Acc})}{P(\text{Acc} \mid A) \cdot P(A)}}\)

Where: \(\small{P(\text{Acc}) = P(\text{Acc} \mid A) \cdot P(A) + P(\text{Acc} \mid B) \cdot P(B)}\)

Thus,

\(\small{P(\text{Acc}) = (0.85 \cdot 0.8) + (0.65 \cdot 0.2) = 0.68 + 0.13 = 0.81}\)

\(\small{P(A \mid \text{Acc}) = \dfrac{0.85 \cdot 0.8}{0.81} = \dfrac{0.68}{0.81} = \dfrac{68}{81}}\)

# Given data
p_A = 80//100
p_B = 20//100

p_acc_given_A = 85//100
p_acc_given_B = 65//100

# Total probability of accuracy
p_acc = p_acc_given_A * p_A + p_acc_given_B * p_B

# Bayes' theorem: P(A | Accurate)
p_A_given_acc = (p_acc_given_A * p_A) // p_acc

println("P(Dept A | Accurate) = $p_A_given_acc")
P(Dept A | Accurate) = 68//81

Q14. A brother and sister apply for two positions in a healthcare research project. The probability of the brother being selected is 1/7, and the probability of the sister being selected is 1/5. What is the probability that:

Answer

Mutually Exclusive Events:

\(\small{P(A \cup B) = P(A) + P(B)}\); \(\small{{P(A \cup B) = \text{A or B (either or both occur)}}}\)

Multiplication Theorem for Independent Events:

\(\small{P(A \cap B) = P(A) \times P(B)}\); \(\small{P(A \cap B) = \text{A or B (both occur)}}\)

Note these formulas for the following questions.

Given:

# Given probabilities
P_B = 1 // 7
P_S = 1 // 5
P_not_B = 6 // 7
P_not_S = 4 // 5
P_both_selected = P_B * P_S

println("P(Both of them will be selected): $P_both_selected")
P(Both of them will be selected): 1//35
P_only_one_selected = P_B * P_not_S + P_not_B * P_S

println("P(Only one of them will be selected): $P_only_one_selected")
P(Only one of them will be selected): 2//7
P_none_selected = P_not_B * P_not_S

println("P(None of them will be selected): $P_none_selected")
P(None of them will be selected): 24//35
P_at_least_one_selected = 1 - P_none_selected

println("P(At least one of them will be selected): $P_at_least_one_selected")
P(At least one of them will be selected): 11//35
Event Probability
Both selected \(\small{\dfrac{1}{35}}\)
Only one selected \(\small{\dfrac{2}{7}}\)
None selected \(\small{\dfrac{24}{35}}\)
At least one selected \(\small{\dfrac{11}{35}}\)